Chapter 6-A

Chapter 6 MATERIAL BALANCE EQUATIONS When a volume of oil is produced from a reservoir, the space originally occupied by the produced oil volume must now be occupied by something else. Unless fluid is injected, the production of oil must result in a decline in reservoir pressure. The decline in reservoir pressure can cause; — influx of fluid from an adjoining aquifer or gas cap. — expansion of fluids in the reservoir. — expansion of reservoir rock grains - reduction of pore volume. There is therefore a relationship between the rate of production and the rate of decline of reservoir pressure. Material balance equations express this dependence in mathematical form. The material balance equation is the basic reservoir engineering analysis tool used to examine past reservoir performance and to predict future performance.

The diagrams in this section are best viewed on the electronic version of the course notes. You may wish to make colour prints of these figures.

134

GAS CAP

OIL ZONE

AQUIFER

Figure 6.1: Schematic of a combination reservoir

6.1

ORIGINAL OIL VOLUME BALANCE

Most material balance equations are written in terms of the original volume occupied by the oil. The volume originally occupied by produced oil may now be occupied by; 1.

expansion of an adjoining gas cap (if one is present).

2.

volume of gas released by oil (if reservoir pressure falls below the bubble point).

3.

volume of oil remaining in the reservoir.

4.

volume occupied by expansion of rock grains - pore space compressibility.

5.

volume occupied by expansion of connate water.

6.

influx of water from an adjoining aquifer (if the aquifer is strong enough).

It should be understood that the attached diagram is for illustrative purposes and is not intended to imply that the materials in the reservoir are segregated in the manner shown. Under some conditions, however, gas released from solution will segregate to the top of the structure. Some of the released solution gas will always be evenly distributed throughout the reservoir, as are the expansions of rock and connate water. 135

Figure 6.2: Material balance on the pore space occupied by the original oil volume

6.1.1

Gas Cap Expansion

If a gas cap is present, and the reservoir pressure drops, the gas cap will expand to replace some of the volume initially occupied by the produced oil. Gas cap expansion = |

where, G Gpc Bg Bgi

= = = =

{z

[RB]

}

(G − Gpc )Bg |

{z

}

−

gas volume at lower pressure

GBgi

| {z }

initial gas volume

original gas cap gas volume, [SCF] cumulative gas production from the gas cap, [SCF] gas formation volume factor at current pressure, [RB/SCF] gas formation volume factor at original reservoir pressure, [RB/SCF]

Note that if Gpc is large, the gas cap may actually shrink rather than expand.Gas cap shrinkage should never be allowed because it always results in a loss of oil recovery. The above equation may be written as, ∆VGCE = (G − Gpc )Bg − GBgi 136

Figure 6.3: Gas cap expansion volume Problem 6.1 - Calculation of gas cap expansion [MBE.mcd] Calculate the change in the size of the gas cap after 20% of the gas cap gas has been produced while the reservoir pressure had dropped from 1225 psig to 1100 psig. The gas cap originally contained 21.3 × 109 SCF. Data: G = 21.3 ×109 Bgc at 1225 psig = 0.002125 Bgc at 1100 psig = 0.002370

[SCF] [RB/SCF] [RB/SCF]

Solution 1. At 1100 psig the gas cap containing 0.8 times the original gas cap volume i.e., G = 0.8Gi 2. ∆VGCE = (G − Gpc )Bg − GBgi

∆VGCE = (0.8 × 21.3 × 109 − 0) 0.002370 − 21.3 × 109 × 0.002125 ∆VGCE = −4.88 × 106 RB

i.e., the gas cap has actually shrunk. This would allow about 5 MMRB oil to migrate into the gas cap, a zone originally free of oil. This would result in a loss of about 1MMRB of oil. We would not allow this to actually happen in the actual reservoir. 137

Problem 6.2 - Calculation of gas cap expansion [MBE.mcd] Calculate the change in gas cap size for the reservoir of Problem 6.1 if the reservoir pressure had dropped from 1225 psig to 900 psig at the time that 20% of the gas was produced. Data: G = 21.3 ×109 Bgc at 1225 psig = 0.002125 Bgc at 900 psig = 0.002905

[SCF] [RB/SCF] [RB/SCF]

Solution 1. At 900 psig the gas cap cumulative production is i.e., Gp = 0.2Gi 2. ∆VGCE = (G − Gpc )Bg − GBgi ∆VGCE = (0.8 × 21.3 × 109 − 0) 0.002905 − 21.3 × 109 × 0.002125 ∆VGCE = 4.24 × 106 RB

i.e., the gas cap has expanded into the oil zone and no oil is lost to the gas zone.

138

Figure 6.4: Released solution gas volume

6.1.2

Released Solution Gas Volume

If the reservoir pressure falls below the bubble point pressure gas will be released from solution. At any time during the production of a reservoir, the gas originally in solution can be placed into three categories; 1.

gas remaining in solution

2.

gas released from solution and produced from the reservoir

3.

gas released from solution but remaining in the reservoir.

On this basis we write, ⎛ ⎜

Released gas volume = ⎜ ⎝ |

{z

[RB]

where, N Np Rsi Rs Bg

= = = = =

}

N R | {z si}

gas originally in solution

⎟ ⎟ Bg ⎠

− (N − Np )Rs −

Gps

gas still in solution

produced solution gas

|

{z

}

|{z}

original oil volume, [STB] cumulative oil produced, [STB] original solution gas-oil ratio, [SCF/STB] solution gas-oil ratio at current pressure, [SCF/STB] gas formation volume factor at current pressure, [RB/SCF]

139

⎞

The above equation is usually written as, ∆VRSG = {N Rsi − (N − Np )Rs − Gps }Bg Problem 6.3 - Calculation of released solution gas volume [MBE.mcd] Cumulative oil production for our example reservoir was 14.73 ×106 STB at the time when reservoir pressure was 900 psig. At the same time cumulative production of solution gas was 4.05 ×109 SCF. Calculate the reservoir volume occupied by released gas. Data: N Rsi at 1225 psig Rs at 900 psig Bg at 900 psig

= = = =

90.46 ×106 230 169 0.002905

[STB] [SCF/STB] [SCF/STB] [RB/SCF]

Solution ∆VRSG = {N Rsi − (N − Np )Rs − Gps }Bg ∆VRSG = {(90.46 × 230 − (90.46 − 14.73) × 169 − 4050) × 106 } 0.002905 ∆VRSG = 11.5 × 106 RB

140

Figure 6.5: Remaining oil volume

6.1.3

Remaining Oil Volume

The oil volume remaining in the reservoir is simply, Reservoir oil volume} = (N − Np )Bo | {z |

[RB]

{z

}

oil still in reservoir

where, N Np Bo

= original oil volume, [STB] = cumulative oil produced, [STB] = oil formation volume factor at current pressure, [RB/STB]

The above equation is written as, ∆VROV = (N − Np )Bo

141

Problem 6.4 - Calculation of remaining oil volume [MBE.mcd] What is the remaining reservoir oil volume for the previous example at 900 psig. Data: Bo at 900 psig = 1.104 [RB/STB]

Solution ∆VROV = (N − Np )Bo ∆VROV = (90.46 × 106 − 14.73 × 106 ) 1.104 ∆VROV = 83.63 × 106 RB

142

Figure 6.6: Rock and connate water expansion volume

6.1.4

Rock and Connate Water Expansion

The expansion of rock and connate water are combined into one term and expressed as the formation compressibility cf . cf is defined as the fractional change in pore volume per psi change in reservoir pressure. The pore volume can be expressed in terms of the original oil volume, as is required for material balance calculations. original oil volume = N Boi = Vp Soi = Vp (1 − Swi ) where, N Boi Vp Soi Swi

= = = = =

original oil volume, [STB] oil formation volume factor at initial pressure, [RB/STB] reservoir pore volume, [RB] initial oil saturation initial or connate water saturation

or, Vp =

µ

N Boi 1 − Swi

143

¶

The total amount of rock expansion is calculated by utilizing the definition of compressibility; 1 ∆V c= V ∆P from which we have ∆V = cV ∆P Rock expansion = formation compressibility × pore volume × pressure change

|

{z

}

[RB]

Rock expansion = cf |

where, cf p1 p

{z

}

[RB]

µ

¶

N Boi (pi − p) 1 − Swi

= formation compressibility, [vol/vol/psi] = initial reservoir pressure, [psig] = current reservoir pressure, [psig]

The total amount of connate water expansion is given by, Water expansion = water compressibility × water volume × pressure change

|

{z

[RB]

}

The initial connate water volume is given by, original connate water volume = Swi Vp = Swi where, cw Swi

µ

N Boi 1 − Swi

¶

= water compressibility, [vol/vol/psi] = initial water saturation

The connate water expansion is, Water expansion = cw Swi

|

{z

[RB]

}

µ

¶

N Boi (pi − p) 1 − Swi

Combining the above equations, we have for rock and water expansion, ∆VRW E

µ

¶

N Boi = (cf + cw Swi ) (pi − p) 1 − Swi

For reservoirs where a gas phase is present, the rock and water expansion term is so small that it may safely be neglected. The rock and connate water term is usually only important for oil reservoirs when the oil pressure remains above the bubble point. 144

Problem 6.5 - Calculation of rock and connate water expansion volume MBE.mcd] What is the rock and water expansion volume for the previous example when the pressure falls from 1225 psig to 900 psig. Data: Swi cf cw

= 0.205 = 3.0×10−6 = 3.0×10−6

[psi−1 ] [psi−1 ]

Solution ∆VRW E

µ

¶

N Boi = (cf + cw Swi ) (pi − p) 1 − Swi

∆VRW E = (cf + cw Swi ) (Vp ) (pi − p) ³

´

∆VRW E = (3.0 + 3.0(0.205)) 10−6 127.84(106 ) (1225 − 900) ∆VRW E = 0.15 × 106 RB

We see that rock and water expansion amounts to about 1% of the released gas volume. This is usually the case in practice and rock and connate water expansion can be neglected for calculations below the bubble point.

145

Figure 6.7: Water influx volume

6.1.5

Water Influx

Unlike the above items in the material balance equation, the volume of water influx cannot be calculated directly. This is because the calculation of water influx requires information characterizing the size and strength of the aquifer and this information is not generally known with any certainty during the early production life of a reservoir. However, since we can calculate all the other terms in the material balance equation, we can determine the net water influx into the reservoir — total water entering the reservoir less water produced — by difference. |Net water {z influx} = [RB]

where, We Wp Bw

W

e |{z}

cumulative water influx

−

Wp Bw

| {z }

cumulative produced water

= cumulative water influx, [RB] = cumulative water produced, [STB] = water formation volume factor at current pressure, [RB/STB]

The above equation may be written as, ∆VN W I = We − Wp Bw

146

6.1.6

General Material Balance Equation

Having analyzed all the individual terms in the material balance equation, we can write, Original oil volume = Gas cap expansion + Released gas volume + Oil volume + Rock and water expansion + Net water influx

∆VOOIP = ∆VGCE + ∆VRGV + ∆VROV + ∆VRW E + ∆VN W I

N Boi = (G − Gpc )Bg − GBgi + {N Rsi − (N − Np )Rs − Gps }Bg + µ ¶ N Boi (N − Np )Bo + (cf + Swi cw ) (pi − p) + 1 − Swi (6.1) We − Wp Bw Note that the two gas production terms are additive and equal to the total gas production at the surface ie., (Gpc + Gps )Bg = Gp Bg The material balance equation may be re-written as, N Boi = G(Bg − Bgi ) + {N Rsi − (N − Np )Rs }Bg − Gp Bg + µ ¶ N Boi (N − Np )Bo + (cf + Swi cw ) (pi − p) + 1 − Swi We − Wp Bw

(6.2)

One of the most important uses of the material balance equation is to calculate water influx into a reservoir. This only requires that we know the volume of initial fluids in-place, reservoir pressure and cumulative oil, gas and water production. Cumulative water influx is, We

µ

N Boi = N Boi − (N − Np )Bo − (cf + Swi cw ) (pi − p) 1 − Swi −GBg + GBgi + Gp Bg − {N Rsi − (N − Np )Rs }Bg +Wp Bw

147

¶

(6.3)

Problem 6.6 - Calculation of water influx from material balance [MBE.mcd] Using the results of the Problems 6.2-5, find the water influx when the pressure falls from 1225 psig to 900 psig. Cumulative water production at this time is 620,000 STB. Data: Wp Bw

= 0.62×106 = 1.0

[STB] [RB/STB]

Solution The following items were calculated in the previous examples. Ex. 2 3 4 5

Item Calculated Gas Cap Exp. Released Gas Remaining Oil Rock & Water Exp.

(G − Gpc )Bg − GBgi {N Rsi − (N − Np )Rs − Gps }Bg (N − Np )Bo ³ ´ N Boi (cf + Swi cw ) 1−S (p − p) i wi

×106 RB 4.24 11.50 83.63 0.15

Re-arranging the material balance equation to calculate We ,

We

µ

N Boi = N Boi − (N − Np )Bo − (cf + Swi cw ) (pi − p) 1 − Swi −GBg + GBgi + Gp Bg − {N Rsi − (N − Np )Rs }Bg +Wp Bw

¶

(6.4)

We = (101.64 − 83.63 − 4.24 − 11.50 − 0.15 + 0.62 × 1.0) × 106 = 2.74 × 106 RB

148

6.2

PRIMARY RECOVERY MECHANISMS

Each of the terms of the material balance equation represents a drive mechanism which contributes to the total energy required to produce oil from the reservoir. If we divide the overall material balance equation by N Boi we obtain,

1 =

(G − Gp )Bg − GBgi {N Rsi − (N − Np )Rs − Gps }Bg + + N B N B oi oi | {z } | {z } Igc

µ

Isg

¶

1 (N − Np )Bo + (cf + Swi cw ) (pi − p) + N Boi 1 − S wi {z } | Ipd

We − Wp Bw N{z Boi } |

(6.5)

Iw

The above equation may be written as, 1 = Igc + Isg + Ipd + Iw where, Igc Isg Ipd Iw

= = = =

gas cap drive index solution gas drive index pressure depletion drive index water drive index

The above indices represent the effectiveness of each drive mechanism for a particular reservoir. A reservoir for which the dominant drive mechanism is, Iw , is called a water drive reservoir. A reservoir for which the dominant drive mechanism is, Igc , is called a gas cap drive reservoir. A reservoir for which the dominant drive mechanism is, Isg , is called a solution gas drive reservoir. 149

Figure 6.8: Reservoir drive index as a function of time

A reservoir for which the dominant drive mechanism is, Ipd , is called a depletion drive reservoir. A reservoir for which more than one drive mechanism is important, is called a combination drive reservoir.

6.2.1

Typical Performance Characteristics for the Different Drive Mechanisms

The dominant drive mechanism for a particular reservoir can often be deduced from the rate of pressure decline and the trend of the producing gas-oil ratio. The trends which are observed when one of the mechanisms dominates are summarized in the following figures. Ultimate recovery is strongly influenced by the type of drive mechanism. 1.

Pressure depletion drive (fluid and rock expansion) results in a rapid straight line pressure response and may be expected to recover less than 5% of the original oil-in-place.

2.

Solution gas drive — pressure drops slowly at first. As the producing GOR increases, pressure falls rapidly. Ultimate recoveries are in the range 15-30 %. 150

Figure 6.9: Reservoir pressure as a function of cumulative oil production

Figure 6.10: Producing GOR as a function of cumulative oil production

151

Figure 6.11: Producing characteristics of gas cap drive reservoirs

3.

Water drive — pressure declines rapidly at first, but the decline rate decreases as water influx from the aquifer increases. The producing GOR remains approximately constant and ultimate recoveries are 50% or more.

152

Figure 6.12: Producing characteristics of solution gas drive reservoirs

Figure 6.13: Producing characteristics of water drive reservoirs

153

6.3

6.3.1

USING MATERIAL BALANCE EQUATIONS Average Reservoir Pressure

The material balance equation describes the whole reservoir in terms of average reservoir pressure, initial volumes of oil, gas and water in-place and cumulative oil, water and gas production volumes. In order to use the material balance equation it is necessary to determine average reservoir pressure for the time (cumulative oil, water and gas production) when the material balance equation is to be applied. Direct measurement of average reservoir pressure would require that all the wells are shut-in and that the bottom-hole pressure is measured at a time after shut-in which is sufficiently long for all the pressure gradients in the reservoir to equalize. This may take months to years (depending on reservoir permeability) and result in considerable loss in revenue because of lost production. The direct measurement approach is therefore impractical. Pressure data obtained from a pressure buildup test can be used to estimate the average pressure in the volume or area drained by the tested well. These tests require short test times (hours to days) and allow reservoir pressure to be mapped over the field. The mapped pressures may be averaged to give the average reservoir pressure as, P pj Vp j p= P (6.6) Vp j

where pj and Vp j are the drainage area pressure and pore volume drained by well j.

6.3.2

Knowns and Unknowns

When solving the material balance equation the parameters may be classified into the following groups of knowns and unknowns. knowns unknowns Np N G Gp Wp We Swc p (Bo , Bg , Rs ) cf cw Bw 154

The PVT properties (Bo , Bg , Rs ) may be considered to be known if the average reservoir pressure is known and representative fluid samples have obtained and analysed. This reduces the number of unknowns to five but we have only one equation. This is the central problem with the use of the material balance equation. Before considering specific examples of the application of the material balance equation, lets consider some of the parameters in the material balance equation. Knowns Np - this is usually known because oil is the primary production target which is sold to generate revenue. Gp , Wp - cumulative gas and water production be unknown for older oil and gas fields because they had little or no value at the time of production. When these are unknown it is not possible to perform material balance calculations for the reservoir. Swc - is usually considered to be accurately known from petrophysical evaluation. cw , Bw - these are known, or may be estimated, from laboratory tests on formation brine. Unknowns N, G - volumetric estimates of the original oil and gas-cap volumes in-place are always known from the field appraisal stage. These are disregarded as soon as production-pressure data becomes available and an attempt is made to estimate in-place volumes from material balance calculations. This is because volumetric estimates include all the mapped hydrocarbon volume, whereas material balance calculation provides the effective volume or the volume which contributes to actual production. This will usually be smaller than the volumetric estimate due to compartmentilization of the reservoir by faults or low permeability zones. We - this is probably the greatest unknown in reservoir development - whether there has been any water influx or not. One of the most important uses of material balance calculations is to estimate water influx. p - although we usually consider reservoir pressure to be known, problems with the interpretation of pressure buildup test data may introduce serious errors and there may be considerable uncertainty in estimates of average reservoir pressure. cf - pore volume or formation compressibility is usually considered to be small and constant. In some cases it may be large and variable. When it is large, compaction may form a major part of the reservoir drive energy and usually leads

155

to considerable subsidence at the surface. This may be of little consequence on land in remote areas but may cause serious problems for operations offshore.

Problem 6.7 Calculating water influx from material balance for history matching aquifer performance [MBE.mcd] The first step in characterizing or history matching an aquifer is to calculate water influx from available production and pressure data. You will be performing this history match in the reservoir engineering course which follows from the present course. After the aquifer has been characterized, the resultant aquifer model is attached to a numerical reservoir simulation model which is used to predict future reservoir performance. The reservoir for this exercise is a small offshore oil field which has been on-stream for only 700 days. The field was shut-in for a 250 day period for upgrading of the gathering system and production facilities. There was a strong and clear aquifer response to the shut-in (use Mathcad to plot the field pressure as a function of time to see the response). The field production history is given in the table below. PRODUCTION HISTORY Time P Np Gp (day) (psia) (MMSTB) (BSCF) 0 50 100 150 200 250 300 350 400 450 500 550 600 650 700

4217 3915 3725 3570 3450 3390 3365 3600 3740 3850 3900 3920 3720 3650 3640

0.00000 0.34270 0.63405 1.06898 1.36788 1.64397 1.91465 1.91465 1.91465 1.91465 1.91465 1.91465 2.11402 2.24921 2.33133

156

0.00000 0.17821 0.32970 0.55587 0.71130 0.85487 0.99562 0.99562 0.99562 0.99562 0.99562 0.99562 1.09929 1.16959 1.21229

Wp (MMSTB) 0.00000 0.00000 0.00121 0.02636 0.07041 0.12218 0.17304 0.17304 0.17304 0.17304 0.17304 0.17304 0.20368 0.25094 0.31690

The initial oil in-place is estimated (volumetrically) to be 33.2 MMSTB. Reservoir system properties were estimated to be: cf = 5×10−6 psi−1 cw = 3×10−6 psi−1 Swc = 0.248 PVT data for the reservoir oil is given in the following table. PVT DATA P Bo Rs Bg Bw (psia) (RB/STB) (SCF/STB) (RB/SCF) (RB/STB) 3365 3465 3565 3665 3765 3865 3965 4065 4165 4265

1.2511 1.2497 1.2483 1.2469 1.2455 1.2441 1.2427 1.2413 1.2399 1.2385

510 510 510 510 510 510 510 510 510 510

-

1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0

Reservoir temperature is 210o F and the gas gravity is 0.69. Determine the cumulative water influx for each of the 50 day time periods in the production history table. Solution hint You need to calculate a cumulative water influx volume for each 50 day period (a total of 14 calculations). Each calculation is identical to the calculations performed in Problems 6-1 to 6-6. You can simply use [MBE.mcd] to do the calculation for each period. Alternatively, you could modify [MBE.mcd] to readin the production data and do the 14 calculations in one pass or prepare your owm spreadsheet using whatever software you are most comfortable with. [Answer: We at 700 days is 2.778 MMRB]

157

6.4

MATERIAL BALANCE FOR A CLOSED OIL RESERVOIR

When water influx from the aquifer is small or negligible (small ineffective aquifer, weak to moderate aquifers during early stages of production when the aquifer response is small) there is negligible water production and the material balance equation reduces to, µ

¶

N Boi 0 = N Boi − (N − Np )Bo − (cf + Swi cw ) (pi − p) 1 − Swi −GBg + GBgi + Gp Bg − {N Rsi − (N − Np )Rs }Bg

(6.7)

If the original gas cap volume may be estimated from a volumetric calculation, the formation compressibility is known, and reliable production-pressure data is available, the only unknown is the original oil in-place, N . This provides a very valuable check on the volumetrically determined value. As discussed previously, the material balance derived value of N may be more reliable than the volumetrically determined volume.

Prediction of future reservoir pressure After the original oil in-place has been established, the material balance equation may be used to predict future reservoir pressure if future cumulative production can be estimated. The main difficulty with this procedure is estimating the volumes of gas and water which will accompany the specified oil production volume. If all the produced oil and water is reinjected, this difficulty is eliminated.

Calculation of original oil in-place by material balance Another important use for the material balance equation for a closed reservoir is to calculate the original oil in-place. This provides an independent check on the volumetric estimate because it does not rely on maps, logs or pore volume estimates. Calculation of original hydrocarbon volume in-place by material balance is the preferred method for gas reservoirs where the method reduces to a simple graphical representation called the p/z-plot. We will demonstrate this method in a following section of the course.

158

Problem 6.8 - Calculation of original oil in-place using material balance equation [OIPMBE.mcd] The discovery pressure of a reservoir containing a gas cap was 3330 psia. Geological evidence suggests that aquifer quality is poor and that water influx into the reservoir is likely to be negligible. The reservoir was produced until the reservoir fell to 2700 psia. Cumulative oil production at this time is 11.503 MMSTB and cumulative gas production is 14.206 BSCF. The volumetric estimates of the original oil-in-place, (N ), is 105 MMSTB and the initial gas cap volume is 81.14 BSCF. Does the material balance calculation confirm this estimate? Data: Swi cf cw

= 0.21 = 3.0×10−6 = 3.0×10−6

[psi−1 ] [psi−1 ]

[Answer: 113 MMSTB]

Solution hints The material balance equation is explicit in N , so we could rearrange it and solve directly for N - you can do this if you like. An alternative way is to solve the material balance equation by trial and error ie., guess a value of N which makes both sides of the material balance equation equal. If you use [OIPMBE.mcd] you will see the I have simply rearranged [MBE.mcd] to allow me to enter a value of N and see the sum of the drive indices with the drive index for water set to zero (closed reservoir condition). The value of N which makes the drive indices sum to one is the correct oil-in-place.

159

6.5

MATERIAL BALANCE FOR A CLOSED GAS RESERVOIR

The material balance equation for a closed gas reservoir is so simple that it may be solved graphically. The graphical solution method is commonly referred to as the P/z-plot. The plot can be used to estimate the original gas in-place and to predict future reservoir pressure given a production forecast. The general material balance equation for a closed gas reservoir reduces to; GBgi = (G − Gpc )Bg Since Bg is given by, Bg = 5.02 × 10−3

zT P

we can write for the above material balance equation (where reservoir temperature is assumed to remain constant), µ

G 5.02 × 10−3

zi T Pi

¶

µ

= (G − Gp ) 5.02 × 10−3

zT P

¶

which may be rearranged to, G or,

µ

P G z or,

zi z = (G − Gp ) Pi P ¶ µ

µ

¶

µ

¶

P = (G − Gp ) z

P P = z z

¶

1 P − Gp G z i

i

i

The above equation results in a straight line relationship between (P/z) and Gp . The straight line on a (P/z)—Gp plot passes through (P/z)i at Gp = 0 and through G at (P/z) = 0. The straight-line relationship is very useful in estimating the initial volume of gas-in-place (G) from limited production history.

160

Figure 6.14: p/z plot for a closed gas reservoir

Problem 6.9 - Calculation of initial gas in-place for a closed gas reservoir [GIP.mcd] Estimate the initial gas content, G, for a closed gas reservoir having the following production history. Pressure Cumulative Gas Gas Deviation (psig) Production (MMSCF) Factor (z) 3500 3350 3200 3050 2750

0 46.7 125.0 203.5 380.0

161

0.84 0.82 0.81 0.80 0.78

Solution

1. Calculate P/z for each pressure point and plot against cumulative production, Gp . Pressure Cumulative Gas Gas Deviation (psig) Production (MMSCF) Factor (z) 3500 3350 3200 3050 2750

0 46.7 125.0 203.5 380.0

0.84 0.82 0.81 0.80 0.78

P/z (psig) 4166.7 4085.3 3950.6 3812.5 3525.6

2. Draw the best-fit straight line through the data and extrapolate to P/z = 0. The intercept at P/z = 0 is the initial gas in place, G. To see this plot open [GIP.mcd]. G = 2.4 BSCF.

162

Figure 6.15: Material balance for a gas reservoir

6.5.1

Water Drive Gas Reservoirs

When a gas reservoir is in contact with a significant aquifer, water influx will occur as the reservoir pressure declines with production. The material balance for such a reservoir may be written as, GBgi = (G − Gp )Bg + We − Wp Bw This equation can be solved for water influx, We = GBgi − (G − Gp )Bg + Wp Bw If G can be evaluated volumetrically and if the field production data is known (Gp and Wp ) at corresponding reservoir pressures, the material balance equation can be used to calculate the cumulative water influx at these times.

163

Figure 6.16: p/z plot for a gas reservoir experiencing water influx

Problem 6.10 - Production history of a water drive gas reservoir [GIP.mcd] The following pressure production data were recorded for a gas reservoir. Can we estimate G from this data? Pressure Cumulative Gas Gas Deviation (psig) Production (MMSCF) Factor (z) 3500 3350 3200 3050 2950

0 52 155 425 880

0.84 0.82 0.81 0.80 0.79

[Answer: yes, G= 2.67 BSCF]

Solution hint Even if the aquifer is active, it takes time for the aquifer to respond to the decrease in reservoir pressure. This means that early in the life of the reservoir will behaive like a closed system. The early data should therefore plot as a straight line on a 164

P/z-plot. Extrapolating this plot to zero pressure will therefore give a realistic estimate of the original gas-in-place. I have setup the spreadsheet [GIP.mcd] to allow you to select the points through which to fit the straight line.

Problem 6.11 - Calculation of water influx for a gas reservoir [GWMTB.mcd] For the water drive gas reservoir of the previous example calculate the cumulative water influx corresponding to the times at which cumulative gas production is given. Cumulative water production is zero for the period and the reservoir temperature is TR = 275 o F. [Answer: 0, 0, 0.016, 0.192, 0.569 MMRB]

Solution hint Do this any way you like. I used [GIP.mcd] from the previous example and added a calculation for Bg from z and the material balance equation for a water drive gas reservoir.

165